Map $z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it },t\in\mathbb{R}$ maps the real axis $(-\infty,\infty)$ onto the unit circle $\mid z\mid=1$. 
Show that the mapping
$$
z^2 = \frac{\frac{1}{2}+it}{\frac{1}{2}-it}, \quad t\in\mathbb{R}
$$
maps the real axis $(-\infty,\infty)$ to the unit circle $|z|=1$.

My try-
$$z^2=\frac{\frac{1}{2}+it}{ \frac{1}{2}-it  }$$
$$|z^2|=\frac{|\frac{1}{2}+it|}{ |\frac{1}{2}-it|  }$$
$$|z^2|=\frac{\sqrt{1/4+t^2}}{  \sqrt{1/4+t^2} }$$
$$|z^2|=1$$
$$|z|^2=1$$
$$|z|=1$$...
 A: The map is multivalued as $z^2=1$ at $t=0$ gives $z=\pm 1$ lying on unit circle $\lvert z\vert=1$.
Added 1-The partial proof you have done shows that the map is onto. But it cannot be a bijection as it is not a single valued function.
Added 2- For another proof, note that
$t=\frac{z^2-1}{2i(z^2+1)}$, where $z=x+iy, x,y\in\mathbb R$
After splitting the RHS into real and imaginary parts, you can equate the imaginary part of $RHS=0$ (Why?) to get the result $x^2+y^2=1$.
A: You argument is partially correct. You did show that $f$ maps $\mathbb{R}$ into a subset of the unit circle, but the problem requires to show the entire unit circle is covered by the image. In other words, if you look at $f:\mathbb{R} \to D_1(0)$, it would be onto (surjective).
To do this, let $z \in D_1(0)$ and show that there is $t \in \mathbb{R}$ such that $f(t)=z$.
A: Let the unit circle be $z=e^{i\theta}$, $\theta\in[0,2\pi]$. Then,
$$t= \frac1{2i} \frac{z^2-1}{z^2+1} = \frac1{2i} \frac{e^{i 2\theta}-1}{e^{i 2\theta} +1} =\frac12\tan\theta
$$
which reveals that $\theta\in[0,\pi)$ and $\theta\in[\pi,2\pi)$ separately map to $t\in(-\infty,\infty)$, thus, not bijective.
